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Standard XII

Physics

Evaporation

For a mixture...

Question

For a mixture of two volatile, completely miscible liquids A and B, with $P_{A}^{0} = 500$ torr and $P_{B}^{0} = 800$ torr. What is the composition of last of liquid remaining in equilibrium with vapour?
Provided the initial ideal solution has a composition of $x_{A} = 0.6$ and $x_{B} = 0.4$ , where $x_{A} and x_{B}$ are the mole fraction of $A$ and $B$ respectively

x_{B} = 0.7

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x_{B} = 0.3

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x_{A} = 0.7

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x_{A} = 0.3

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Solution

The correct option is C $x_{A} = 0.7$
Given,
$P_{A}^{0} = 500 torr; P_{B}^{0} = 800 torr$

Let,
mole fracion of $A$ in vapor = $y_{A}$
mole fracion of $B$ in vapor = $y_{B}$
When most of the liquid has vapourized $y_{A}$ would be, $0.6$ since $x_{A}$ is $0.6$ (Given)
$y_{A} = \frac{p_{A}^{0} x_{A}}{P_{A}^{0} x_{A} + P_{B}^{0} x_{B}} = \frac{p_{A}^{0} x_{A}}{P_{A}^{0} x_{A} + P_{B}^{0} (1 - x_{A})} \Rightarrow 0.6 = \frac{500 x_{A}}{500 x_{A} + 800 (1 - x_{A})} \Rightarrow - 180 x_{A} + 480 = 500 x_{A}$
$\Rightarrow x_{A} \approx 0.7$
$∴ x_{B} = 1 - x_{A} = 0.3$

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